③
13??21P?x＜１,y＞?=??dxdy??10?3/2dxdy?1/4
22?D2?7. 证明：P?x1＜X?x2?P?y1＜Y?y2???Fx?x2??Fx?x1??Fy?y2??Fy?y1? ?Fx,y?x2,y2??Fx,y?x2,y1??Fx,y?x1,y2??Fy?x1,y1? ?P?x1＜X?x2,y1＜Y?y2? 故独立得证. 8. ??x1|x2?1 ０ 1 3１ 2 3P x９. 解：① P ②Py?yj?ij?PiPji???i?1pipji ③Pij?xpipjix?ipipjix
111P?P??0 P 1322124421??p?y??1?p?x?0? 故不独立. ② P11410. 解：① P11?11. x y y1 y2 y3 pix x1 1/24 1/8 1/8 3/8 1/2 1/12 1/4 1/3 p1,p2,?,pn1/4 3/4 1 x2 pjy1/6 p11,p12,?,p1np21,p22,?,p2n独立p1,p2,?,pn?pp,?p１２m12. 证明：必要性： 由：
??????pm1,pm2,?,pmn?秩为1
p1,p2,?,pnp11,p12,?,p1np11,p12,?,p1np21,p22,?,p2n秩为1p21,p22,?,p2n?充分性，若
??????pm1,pm2,?,pmnpm1,pm2,?,pnm 26
?p?x?x2y?y1??p?x?x2y?y2??p?x?x2y?y3???p?x?x2y?yn??p?x?xy?y??p?x?xy?y??p?x?xy?y???p?x?xy?y??1112131n??
.............??p?x?xmy?y1??p?x?xmy?y2???p?x?xmy?yn?? 从上式可得x与y独立. 13. 解：① X2?Bx?c?0,
有实根的概率 B2?4c?0,PB2?4C?0
???P?2,1??P?3,1??P?3,2??P?4,4??P?4,2??P?4,1??P?4,3?
?P?5,1??P?5,2??P?5,3??P?5,4??P?5,5??P?5,6??P?6,1??P?6,2??P?6,3??P?6,4??P?6,5???P?6,6?
=
19 36② PB?4C?0?P?2,1??P?4,4??2??1 1814. 解：fx1y1?x,y??f?x,y? fy1?y?x＞0 其它k1e?3x?4y?3e?3x?? 当 y?0时， fx1y1?x,y??k1?4y?0e3 当 y＜0时， fx1y1?x|y??0x????,???
?4e?4y,y?0 当 x?0时， fy1|x1(y|x)＝f1(x,y)/fX1(x)??
?0,y?0 当 x＜0时 fy1x1?xy??0??＜y＜??
f2?x,y??3e?3x?3y??② 当 y?0时， fx2y2?x,y??fy2?y??0Y＜0时，fx2x?y x?yy2(x,y)?0 x?(??,??)
?4e?4y0?y?x?f(x,y)?4x??当X≥0时，fy2x2(yx)?2 (1?e)fx2z(x)?0其它?当X＜0时，
fy2x2(yx)＝0,y?(??,??)
215.解 1.由S(D)=4?1/2?7
27
得X与Y的联合密度函数为f(x,y)???27(x,y)?D
其它?0?2/70?x?y?12. 由于0≤y≤1时，f(x,y)??
0其它?从而 0≤y≤1时，fy(y)??????f(x,y)dx??y?102dx?2(y?1)
77
??27又 1≤y≤3时，f(x,y)????0从而 1≤y≤3时，fy(y)?y?1?x?2其它2y?1?????f(x,y)dx??2/7dx?2(3?y); 7又当Y＜0时,Y＞3时，f(x,y)=0,从而fY(y)＝?????f(x,y)dx?0
2y?1)/70?y?1?（?综上得： fy(y)??（23?y)/71?y?3
?0其它???270?y?x?1此外，0≤X≤1时，f(x,y)??
?其它?0从而 0≤X≤1时，fx(x)??????f(x,y)dx??x?102dy?2(x?1)
77
当 1≤X≤2时，f(x,y)??7从而 1≤X≤2时，fx(x)?当X＜0或X＞2时，f(x,y)=0 从而fx(x)???2??0x?1?y?x?1其它x?1?????f(x,y)dy??2dy?4
7x?17?????f(x,y)dy?0
2x?1)/70?x?1?（?41?x?2 综上得：fx(x)??7?0其它?16. 证明：
??1Da?x?b又?（x)?y??(x) （１）f(x,y)??
?0其它?fY(y)??ba?b?a1dx???DD??0a?x?b,?(x)?y??(x)其它
1f(x,y)D当a?x?b,?(x)?y??(x)时,fxy(x|y)??b?afY(y)
28
?D1 b?a1f(x,y)D?fxy(x|y)??b?afY(y)故fxy(x|y)为均匀分布
?1??(x)?y??(x),a?x?b ??b?a其它D??0又 fX(x)????(x)(x)??(x)??(x)1dy???DD?0?a?x?b其它
fyx11?f(x,y),a?x?b,?(x)?y??(x)?D ?????(x)??(x)fX(x)?(x)??(x)?0?D故 fyx(yx)为均匀分布.
17. 解： f?x,y??fy|xyxfx?x? fy?y?????????f?x,y?dx
故fxy?x,y??fyx?yx?f?x??????f?x,y?dx
18. f?x,y??fX?x??fY?y? 故独立. g?x,y??gX?x?gY?y? 不独立. 19. f?x,y??fX?x?fY?y? 不独立. 20. f?x????10?x?1
0其它?2?1?y?e2 f?y???2??0????y???,0?x?1其它
21. 若?x,y?服从二元正态分布，则
f?x,y??12????21??2?x?u1?22?12?e21??2?1???x?u?2(x?u1)(y?u2)?y?u2?2??2????22???2??12??1??1f?x??e2??1
?1f?y??e2??229
?y?u2?22?22
因为 f?x?,f?y?均无参数ρ，故可见，f?x,y?不能由fX?x?,fY?y?决定.
?fX?x??????22. 解：
1e2??x22x2?y2?2?1?sinxsiny?dy
1?1?sinxsiny?dy?e2?y2?2
1?e2??????e?y22?x2?????e?y22dy1?e2?23. x2?21,fY?y??e2? ???y???
fX?x?,fY?y?均服从正态分布，但是f?x,y?不服从正态分布.
? ０ １ ２ ３ ４ P 1/9 4/9 4/9 0 0 ? 0 1 2 -1 -2 P 3/9 2/9 1/9 2/9 1/9 ? ? -2 -1 0 1 2 ０ １ ２ ３ ４ ０ 0 1/9 0 0 0 1/9 0 1/9 0 1/9 0 1/9 0 1/9 0 1/9 0 1/9 0 0 0 1/9 0 0 1/9 2/9 3/9 2/9 1/9 p????i? 1/9 2/9 3/9 2/9 1/9 1 p????j? 24. ? ? 25.证明：
2 3 4 5 6 7 8 9 10 11 12 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 当n=2时，P?X1＋X2＝k???P?Xi?0kk1?i,X2?k?i? ?i,??p?X2?k?i?
??i?k2 ??P?Xi?0k1 ??i?0i?1i!??e??1(k?i)!?ik?i?1?2
?e?(?1??2)?i?ok
i!(k?i)!30